Since $(2k)!!=2^k k!$ the series is the Taylor series expansion of $1/\Gamma((3+p)/2)$ at $p=0$
–
HeikeJan 13 '12 at 9:35

So the answer would be $\frac{1} {{\Gamma \left( {\frac{{3 + p}} {2}} \right)}} - \frac{1} {{\Gamma \left( {\frac{3} {2}} \right)}}$. Thank you, very clever, I'm tired so didn't see it. If you want to make it into the answer, I will accept it and upvote it
–
AlenJan 13 '12 at 9:58

1 Answer
1

By writing $(2k)!!$ as $(2k)!!=2^k k!$ the series becomes
$$
\sum_{k=1}^\infty \frac{1}{k!}\left(\frac{p}{2}\right)^k \left.\frac{d^k}{ds^k}\frac{1}{\Gamma(s)}\right|_{s=\frac{3}{2}}
$$
which is by definition the Taylor series expansion of
$$
\frac{1}{\Gamma\left(\frac{3+p}{2}\right)}-\frac{1}{\Gamma\left(\frac{3}{2}\right)}
$$
at $p=0$.